Contour integral:
If function $f:\Bbb C-\{0\}\to\Bbb C$ continuous function on punctured domain.Suppose anti derivative for $f$ exists everywhere except $\{0\}$ then on a closed contour $C$, which does not pass through origin, $\int_Cf(z)\,dz=0$.
Am I correct above?
If I am correct, then why $\int_C \frac1z\,dz=0$ does not hold? ($\operatorname{Log} z$ is derivative of $1/z$ except at zero, right?)
No, you are not correct, precisely because $\int_C\frac1z\,\mathrm dz\neq0$ for some closed contours. And there is no differentiable function $\operatorname{Log}\colon\mathbb C\setminus\{0\}\longrightarrow\mathbb C$ such that $(\forall z\in\mathbb C\setminus\{0\}):\operatorname{Log}'(z)=\frac1z$. The map $z\mapsto\frac1z$ only has a antiderivative locally, not globally.